Proof of Bernoulli's theorem
Consider a fluid of negligible viscosity moving with laminar flow, as shown in Figure 1.
Let the velocity, pressure and area of the fluid column be v1, P1 and A1 at Q and v2, P2 and A2 at R. Let the volume bounded by Q and R move to S and T where QS = L1, and RT = L2. If the fluid is incompressible:
A1L1 = A2L2
The work done by the pressure difference per unit volume = gain in k.e. per unit volume + gain in p.e. per unit volume. Now:
Work done = force x distance = p x volume
Net work done per unit volume = P1 - P2
k.e. per unit volume = ½ mv2 = ½ Vρ v2 = ½ρv2 (V = 1 for unit volume)
k.e. gained per unit volume = ½ ρ(v22 - v12)
p.e. gained per unit volume = ρg(h2 – h1)
where h1 and h2 are the heights of Q and R above some reference level. Therefore:
P1 - P2 = ½ ρ(v12 – v22) + ρg(h2 - h1)
P1 + ½ ρv12 + ρgh1 = P2 + ½ ρv22 + rgh2
P + ½ ρv2 + ρgh is a constant
For a horizontal tube h1 = h2 and so we have:
P + ½ ρv2 = a constant
This is Bernoulli's theorem You can see that if there is a increase in velocity there must be a decrease of pressure and vice versa.
No fluid is totally incompressible but in practice the general qualitative assumptions still hold for real fluids.